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A water molecule has an electric dipole moment 6.4 x 10 cm when it is in vapour state. The distance in metre between the centre of positive and negative charge of the molecule is (a) 4x10-10 104 10-12 (b) 4 x 10-11 (d) 4 x 10-13

Question

A water molecule has an electric dipole moment 6.4 x 10 cm when it is in vapour state. The distance in metre between the centre of positive and negative charge of the molecule is (a) 4x10-10 104 10-12 (b) 4 x 10-11 (d) 4 x 10-13

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Solution

The electric dipole moment (p) of a molecule is given by the product of the charge (q) and the distance (d) between the charges. In other words, p = q * d.

The charge of an electron (which is the same as the charge of a proton) is approximately 1.6 x 10^-19 Coulombs.

Given that the dipole moment of a water molecule is 6.4 x 10^-30 C.m, we can rearrange the formula to solve for d:

d = p / q d = (6.4 x 10^-30 C.m) / (1.6 x 10^-19 C) d = 4 x 10^-11 m

So, the distance between the centre of positive and negative charge of the water molecule is 4 x 10^-11 m. Therefore, the correct answer is (b) 4 x 10^-11.

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